Leetcode #1190: Reverse Substrings Between Each Pair of Parentheses
In this guide, we solve Leetcode #1190 Reverse Substrings Between Each Pair of Parentheses in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a string s that consists of lower case English letters and brackets. Reverse the strings in each pair of matching parentheses, starting from the innermost one.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, String
Intuition
The problem has a natural nested or last-in-first-out structure.
A stack lets us resolve matches in the correct order as we scan.
Approach
Push items as they appear and pop when you can finalize a decision.
The stack captures the unresolved part of the input.
Steps:
- Push elements as you scan.
- Pop when a rule or match is satisfied.
- Use the stack to compute results.
Example
Input: s = "(abcd)"
Output: "dcba"
Python Solution
class Solution:
def reverseParentheses(self, s: str) -> str:
stk = []
for c in s:
if c == ")":
t = []
while stk[-1] != "(":
t.append(stk.pop())
stk.pop()
stk.extend(t)
else:
stk.append(c)
return "".join(stk)
Complexity
The time complexity is , and the space complexity is , where is the length of the string . The space complexity is , where is the length of the string .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.